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This Chapter Deals with Conservation of Energy, Momentum and Angular Momentum in Electromagnetic Systems

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This Chapter Deals with Conservation of Energy, Momentum and Angular Momentum in Electromagnetic Systems This chapter deals with conservation of energy, momentum and angular momentum in electromagnetic systems. The basic idea is to use Maxwell’s Eqn. to write the charge and currents entirely in terms of the E and B-fields. For example, the current density can be written in terms of the curl of B and the Maxwell Displacement current or the rate of change of the E-field. We could then write the power density which is E dot J entirely in terms of fields and their time derivatives. We begin with a discussion of Poynting’s Theorem which describes the flow of power out of an electromagnetic system using this approach. We turn next to a discussion of the Maxwell stress tensor which is an elegant way of computing electromagnetic forces. For example, we write the charge density which is part of the electrostatic force density (rho times E) in terms of the divergence of the E-field. The magnetic forces involve current densities which can be written as the fields as just described to complete the electromagnetic force description. Since the force is the rate of change of momentum, the Maxwell stress tensor naturally leads to a discussion of electromagnetic momentum density which is similar in spirit to our previous discussion of electromagnetic energy density. In particular, we find that electromagnetic fields contain an angular momentum which accounts for the angular momentum achieved by charge distributions due to the EMF from collapsing magnetic fields according to Faraday’s law. This clears up a mystery from Physics 435. We will frequently re-visit this chapter since it develops many of our crucial tools we need in electrodynamics. 1 We begin with energy conservation and obtain the Poynting vector which you have had exposure to in Physics 212. Poynting was Maxwell’s graduate student. The Poynting vector S is very similar to the current density J. Just like the current density gives the current per unit area flowing into a region, the Poynting vector gives the electromagnetic intensity or the power per unit area flowing into a region. We start with an element of power which is force dotted into velocity. Later we will call this the “mechanical” power density which is essentially the rate of change of the work done on the charges by the electromagnetic field per unit volume. As you will see, besides this mechanical energy there is also the energy density stored in the electric and magnetic fields themselves. For the “mechanical” power, the force is just the electric and magnetic force times for an element of charge dq dotted into velocity where only the E-field counts since the B-field does no work. The element of charge is rho times an element of volume. J is just rho v and hence the total work done on the charge by the field per unit volume is E dot J. The basic strategy for this chapter is to use Maxwell’s Eqn. to eliminate all sources in favor of field expressions. In this case, we can eliminate J from the Maxwell-Ampere law which gives us one term from Ampere’s law and one term from Maxwell’s displacement current. The displacement current power contribution is then proportional to E dot partial E/partial t, and hence Maxwell term can be written as the time derivative of epsilon_0 E^2/2 which you recognize as the electric field energy density which we will call u_e. The volume of u_e is the total energy stored in the electric field as we discussed in the Potential Chp. The Ampere law contribution which is the B field dotted into the curl of E can be re-written using one of the product rules in Griffiths’ Chp. 1. After this product rule is applied one is left with the time derivative of the magnetic energy u_m and the divergence of E cross B. The quantity E cross B/mu_0 is the Poynting vector S. We can write E dot J as the power density or rate of change of a mechanical energy density u_mech. The rate of change of the integral of u_mech is the total power flowing into the volume. With the u_mech definition, we can write our power expression as the divergence of S plus the rate of change of the sum of all three densities (mechanical, electrical and magnetic) is zero. This is in close analogy with the continuity Eq. that says the divergence of J gives the rate of change of the charge density (free and bound). Just like the surface integral of J is the current flow into or out of a volume, the surface integral of S is the power flowing into or out of a volume. 2 Our coaxial cable where current flows down the inner conductor and back along the outer conductor gives an example of the use of the Poynting Vector. To compute S we need an expression for B and E. We obtained B using Ampere’s law in our treatment of Griffiths Ex 7.13. This B field is confined to a < s < b. To compute E we need to know something about the charges. A simple assumption is the inner conductor has a linear charge density of lambda and the outer conductor has a charge density of – a*lambda/b. This is indeed what nature would set up if we put a voltage V between the two conductors. The current I could then be viewed as 2 pi*a*lambda times v in the same way we defined currents in our Magnetostatics Chp. But since we don’t know the charge velocity we treat lambda and I as independent variables. The same current is returned on the outer conductor so B disappears at s> b. We can then construct E by applying Gauss’s Law using a Gaussian cylinder. The Poynting vector is just the cross product of E and B which points in the z-hat direction or the direction of the inner conductor current. Integrating S over the area of the leftmost cylinder end gives the total power flowing into the coaxial cable. We are using the Gauss’s law “outward” area convention which accounts for the – sign. We get a simple expression involving the current, charge density and the log (b/a). This power should just be the usual VI power. To check this we need to compute the voltage by integrating our E-expression along s from “a” to “b”. This gives us a (hopefully) familiar expression involving lambda and log(b/a). Indeed the product of V and I is the same as the surface integral of the Poynting vector S. Although the power flows into the coax, it also flows out through the other end so the net power flow is zero and the energy inside the cable is constant. The situation is analogous to current flow through a wire. The same current flows into the wire that leaves the wire so the charge within the wire is constant. Since the total surface integral of the Poynting vector vanishes, the divergence theorem tells us that the volume integral of the divergence of S must vanish. In fact the divergence of S is zero for this case since the E and B fields are constant meaning their energy densities are constant and E is perpendicular to J so E dot J vanishes. Indeed the divergence of our S expression is zero. 3 Lets apply the Poynting vector approach to a cylindrical resistor of radius a , length h and conductivity sigma. Recall in a resistor, the E-field is proportional to J and J is a constant vec J = J_0 z-hat. We can find the B-field within the resistor using Ampere’s law and we find that it is proportional to s and points in the phi-hat direction. We then form the cross product of E and B to get a Poynting vector which points in the –s-hat direction. This is a bit surprising since one would naively think the power flows into the ends rather than into the resistor through the cylinder walls. We next compute the total power flow out of the resistor by integrating the Poynting vector. We need to consider the two ends (#1 and #2) and the barrel (#3). Since the end area vectors are in the +/- z-hat direction and the Poynting vector is in the –s-hat direction, only the barrel region contributes. We get a simple integral over d phi for the barrel and a negative power flow out of the resistor indicating that the power flows into the resistor. We all know that the power into a resistor is V I and we can calculate V from V = h E. The current is just J_0 pi a^2. We can insert I and E into our power expression to get exactly the same power that we computed using the Poynting vector. But where does this power go? The electric and magnetic fields are both static so it cannot go into changing U_EM. The power does go into mechanical energy since it equals E dot J d tau The current and voltage are constant so it cannot be increasing the KE + PE of the charges. Evidently the mechanical power goes into heat. This heat presumably radiates radially from the resistor, so perhaps it makes sense that the Poynting vector flows radially into the resistor to compensate for the heat loss. But something doesn’t make sense when we try to compute the power with a cylinder with a radius larger than “a”.
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